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DochEvi [55]
2 years ago
10

In a soda-acid fire extinguisher, concentrated sulfuric acid reacts with sodium hydrogen carbonate to produce carbon dioxide, so

dium sulfate, and water.
a. how many moles of sodium hydrogen carbonate would be needed to react with 150.0 g of sulfuric acid?
b. how many moles of each product would be formed?
Chemistry
1 answer:
vlabodo [156]2 years ago
7 0
Chemical reaction: 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2CO₂ + 2H₂O.
m(H₂SO₄) = 150.0 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 150 g ÷ 98 g/mol.
n(H₂SO₄) = 1.53 mol.
From chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 : 2.
n(NaHCO₃) = 2 · 1.53 mol = 3.06 mol.
n(H₂SO₄) : n(Na₂SO₄) = 1 : 1.
n(Na₂SO₄) = 1.53 mol.
n(H₂SO₄) : n(CO₂) = 1 : 2.
n(CO₂) = 3.06 mol.
n(H₂SO₄) : n(H₂O) = 1 : 2.
n(H₂O) = 3.06 mol.

You might be interested in
Identify the oxidizing and reducing agents in the following: 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l)
schepotkina [342]

Answer :  The oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)

The half oxidation-reduction reactions are:

Oxidation reaction : Fe^{2+}\rightarrow Fe^{3+}+1e^-

Reduction reaction : O^-+1e^-\rightarrow O^{2-}

The oxidation state of oxygen in H_2O_2 and H_2O is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, 'Fe^{2+}' act as a reducing agent and 'H_2O_2' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

7 0
1 year ago
0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.
max2010maxim [7]

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

7 0
2 years ago
Electrons in __________ bonds remain localized between two atoms. electrons in __________ bonds can become delocalized between m
dsp73

Electrons in sigma <span>bonds remain localized between two atoms. Sigma </span><span>bond results from the formation of </span><span>a molecular orbital </span><span>by the end to </span><span>end overlap of atomic </span>orbitals. Electrons<span> in pi</span> bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals. 

 

<span> </span>

7 0
1 year ago
A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)
AnnyKZ [126]

Answer:

\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

                   0.0150          x

K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

                        x          2.293 × 10⁻⁵

K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}

 

4 0
1 year ago
N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t
son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.


Explanation : When nitrogen (N_{2})and hydrogen (H_{2})are mixed, the mole ratio becomes 1 : 1.5,


Now we know that (H_{2}) is acting as a limiting agent.


So at the time of when 0.4 moles of (NH_{3}) is been formed it requires 0.4 moles of (N_{2}) and 3.4 moles of (H_{2})


So, we find the the remaining (N_{2}) will be 0.6 and

(H_{2}) will be 0.3 mole present in mixture.


So, the mole fraction of (N_{2}) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0
2 years ago
Read 2 more answers
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